Problem: What are the first three non-zero terms of the Maclaurin polynomial for the function $f(x)=\dfrac{2}{3}{{e}^{3x}}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\frac{2}{3}+2x+3x^2$ (Choice B) B $2+2x+3x^2$ (Choice C) C $ 1+3x+\frac{9x^2}{2!}$ (Choice D) D $\frac{2}{3}+3x+6x^2$
Solution: Start with the Maclaurin series for $e^x$. $ e^x=1\text{ }+\text{ }x\text{ }+\text{ }\frac{{{x}^{2}}}{2!}\text{ }+\text{ }\frac{{{x}^{3}}}{3!}\text{ }+\ldots +\text{ }\frac{{{x}^{n}}}{n!}\text{ }+\ldots \,$ Substitute $3x$ for $x$. $ e^{3x}=1\text{ }+3x\text{ }+\text{ }\frac{9{{x}^{2}}}{2!}\text{ }+\text{ }\frac{27{{x}^{3}}}{3!}\text{ }+\ldots +\text{ }\frac{{{(3x)}^{n}}}{n!}\text{ }+\ldots$ Multiply by $\dfrac{2}{3}$. $\begin{aligned} &\phantom{=}\frac{2}{3}e^{3x} \\\\ &=\dfrac{2}{3}+2x+\dfrac{6{{x}^{2}}}{2!}+\dfrac{18{{x}^{3}}}{3!}+\ldots +\dfrac{2{{(3x)}^{n}}}{3n!}\text{ }+\ldots \\\\ &=\dfrac{2}{3}+2x+3x^2+3x^3+\ldots+\dfrac{2{{(3x)}^{n}}}{3n!}+\ldots \end{aligned}$ The first three non-zero terms are $\dfrac{2}{3}+2x+3x^2$.